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    conditional binomial is a poisson distribution




    Hi, I'm fighting with this problem for a while. There is something important I'm missing

    So the problem goes like this X~Poisson(lambda) and Y|X=k ~ Bi(k,p).

    Prove Y~Poisson(lambda*p)

    I already got to show that if Y~Poisson(mu) then X|X+Y=k ~ Bi(k, lamda/(lambda+mu))

    So I to start the problem I thought

    P(Y=y)=sum( P(Y=y|X=k)*P(X=k)) for k>=0

    Eventually I get something like this

    P(Y=y)= (p/1-p)^y * e^(-lamda)/y! * sum ( (lamda-p)^k / (k-n)!)

    I tried to move on and I can't. After long while trying I thought of solving the problem through moment generator functions. But I'm very lost because I got a some of multpilications, is it true that Mt(Y) = sum(Mt(Y|X=k)*Mt(X) ?

    Thanks for reading and any hint

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    Re: conditional binomial is a poisson distribution

    I think directly attacking the pmf is not hard also:

    Given X \sim \mathrm{Poisson}(\lambda),
Y|X = k \sim \mathrm{Binomial}(k, p)

    \Pr\{X = k\} = e^{-\lambda}\frac {\lambda^k} {k!}, k = 0, 1, 2, ..

    \Pr\{Y = y|X = k\} = {k \choose y} p^y (1 - p)^{k - y},
y = 0, 1, 2, ..., k

    For y = 0, 1, 2, ...

    \Pr\{Y = y\} = \sum_{k=y}^{+\infty} \Pr\{Y = y|X = k\}\Pr\{X = k\}

    = \sum_{k=y}^{+\infty} {k \choose y} p^y (1 - p)^{k - y}
e^{-\lambda} \frac {\lambda^k} {k!}

    = e^{-\lambda} \frac {(\lambda p)^y} {y!} \sum_{k=y}^{+\infty}
\frac {[\lambda (1 - p)]^{k-y}} {(k - y)!}

    = e^{-\lambda} \frac {(\lambda p)^y} {y!} \sum_{k=0}^{+\infty}
\frac {[\lambda (1 - p)]^{k}} {k!}

    = e^{-\lambda} \frac {(\lambda p)^y} {y!} e^{\lambda(1 - p)}

    = e^{-\lambda p} \frac {(\lambda p)^y} {y!}

    \Rightarrow Y \sim \mathrm{Poisson}(\lambda p)

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    Re: conditional binomial is a poisson distribution

    Thanks a lot.

    I was trying to do that but without adding the lambda^y * lambda^-y I couldn't get anywhere. I even knew I had to put a lambda^y, but I couldn't find how

    I guess having the k-y! should have been a hint.

    It seems very clear now, thanks again.

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    Re: conditional binomial is a poisson distribution

    Hi,
    I am struggling with this same problem setup: X~ Poi(λ),Y|X=k ~Bin(k,p)

    But I need to show:
    (a) Using the moment generating functions show that Y has a Poi(λp) distribution.
    (b) Show that Y and X-Y are independent and find the conditional distribution of X given Y=y

    I have solved part (a) using mgfs as it directed. I will show my solution since this approach was mentioned above with no solution:

    E(e^tY )=E[E(e^tY│X=k) ]
    = E[(pe^t+(1-p) )^k ]
    = ∑〖(pe^t+(1-p) )^k P(X=k) 〗
    =∑〖(pe^t+(1-p) )^k e^(-λ) λ^k/k!〗
    =∑〖(λpe^t+λ(1-p) )^k e^(-λ)/k!〗
    = e^(-λ) ∑(λpe^t+λ(1-p) )^k/k!
    = e^(-λ) e^(λpe^t+λ(1-p) )
    = e^(λpe^t-λp)=e^(λp(e^t-1))

    Which is the mgf for the Poi(λp) distribution.

    (where the summation is from k=0 to ∞ in each line involving a summation above- Sorry, I am not familiar with how to post equations.)



    Part (b) is where I have run into trouble. I have shown (hopefully correctly) that X-Y~Bin(k,1-p):

    P[X-Y=u]
    =P[Y=k-u|X=k]
    = kC(k-u) p^(k-u) 〖(1-p)〗^u
    = kCu 〖(1-p)〗^u p^(k-u)

    Which is the pgf of the Bin(k, 1-p) distribution.

    I have tried unsuccessfully to show that X and X-Y are independent.

    I have also tried unsuccessfully to find the conditional distribution of X|Y=y.

    I think I’m missing something important because I fail to see the need to any relationship between the independence of X and X-Y and the conditional distribution of X|Y=y while the wording of the problem suggests that these two are somehow related…

    Any help/hints/tips/etc. would be greatly appreciated.

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    Re: conditional binomial is a poisson distribution

    For u = 0, 1, 2, ...,

    \Pr\{X - Y = u\} 
= \sum_{k=u}^{+\infty} \Pr\{X - Y = u|X = k\}\Pr\{X = k\}

    =\sum_{k=u}^{+\infty} \Pr\{Y = k - u|X = k\}\Pr\{X = k\}

    =\sum_{k=u}^{+\infty} {k \choose k - u}p^{k-u}(1 - p)^u 
e^{-\lambda} \frac {\lambda^k} {k!}

    = e^{-\lambda} \frac {[\lambda(1 - p)]^u} {u!} 
\sum_{k=u}^{+\infty} \frac {(\lambda p)^{k-u}} {(k -u)!}

    = e^{-\lambda} \frac {[\lambda(1 - p)]^u} {u!}
\sum_{k=0}^{+\infty} \frac {(\lambda p)^k} {k!}

    = e^{-\lambda} \frac {[\lambda(1 - p)]^u} {u!} e^{\lambda p}
= e^{-\lambda(1 - p)} \frac {[\lambda(1 - p)]^u} {u!}

    \Rightarrow X - Y \sim \mathrm{Poisson}(\lambda(1 - p))

    Then consider

    \Pr\{Y = y, X - Y = u\} 
= \Pr\{Y = y, X - Y = u|X = u + y\}\Pr\{X = u + y\}

    = \Pr\{Y = y|X = u + y\}\Pr\{X = u + y\}

    = {u + y \choose y } p^y (1-p)^u e^{-\lambda} \frac {\lambda^{u+y}} {(u+y)!}

    = e^{-\lambda p} \frac {(\lambda p)^y} {y!}
e^{-\lambda(1 - p)} \frac {[\lambda (1-p)]^u} {u!}

    = \Pr\{Y = y\}\Pr\{X - Y = u\} ~~ \forall u, y

    \Rightarrow Y, X - Y are independent.

    Finally, \Pr\{X = k|Y = y\} = \Pr\{X - y = k - y|Y = y\}
= \Pr\{X - Y = k - y|Y = y\}

    = \Pr\{X - Y = k - y\} 
= e^{-\lambda(1 - p)} \frac {[\lambda(1 - p)]^{k - y}} {(k-y)!}, 
k = y, y+1, y+2, ...

    which is a Poisson distribution shifted by y

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    Re: conditional binomial is a poisson distribution

    Thank you so much BGM! That was such a silly mistake that I was making in finding the distribution of X-Y. Once I have the correct distribution for X-Y showing Y and X-Y are independent is simple. That is a nice trick for finding the conditional distribution. I'm not sure I would have thought of that. Thanks for your help!

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    Re: conditional binomial is a poisson distribution

    Quote Originally Posted by BGM View Post
    Then consider

    \Pr\{Y = y, X - Y = u\} 
= \Pr\{Y = y, X - Y = u|X = u + y\}\Pr\{X = u + y\}
    I wonder a bit about this; why isn't it:
    \Pr\{Y = y, X - Y = u\} 
=\sum_{u+y=0}^{+\infty} \Pr\{Y = y, X - Y = u|X = u + y\}\Pr\{X = u + y\}

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    Re: conditional binomial is a poisson distribution

    Yes that is my silly mistake. It should be

    \Pr\{Y = y, X - Y = u\} = \Pr\{Y = y, X = u + y\}

    instead.

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    Re: conditional binomial is a poisson distribution

    Quote Originally Posted by BGM View Post
    I think directly attacking the pmf is not hard also:

    Given X \sim \mathrm{Poisson}(\lambda),
Y|X = k \sim \mathrm{Binomial}(k, p)

    \Pr\{X = k\} = e^{-\lambda}\frac {\lambda^k} {k!}, k = 0, 1, 2, ..

    \Pr\{Y = y|X = k\} = {k \choose y} p^y (1 - p)^{k - y},
y = 0, 1, 2, ..., k

    For y = 0, 1, 2, ...

    \Pr\{Y = y\} = \sum_{k=y}^{+\infty} \Pr\{Y = y|X = k\}\Pr\{X = k\}

    = \sum_{k=y}^{+\infty} {k \choose y} p^y (1 - p)^{k - y}
e^{-\lambda} \frac {\lambda^k} {k!}

    = e^{-\lambda} \frac {(\lambda p)^y} {y!} \sum_{k=y}^{+\infty}
\frac {[\lambda (1 - p)]^{k-y}} {(k - y)!}

    = e^{-\lambda} \frac {(\lambda p)^y} {y!} \sum_{k=0}^{+\infty}
\frac {[\lambda (1 - p)]^{k}} {k!}

    = e^{-\lambda} \frac {(\lambda p)^y} {y!} e^{\lambda(1 - p)}

    = e^{-\lambda p} \frac {(\lambda p)^y} {y!}

    \Rightarrow Y \sim \mathrm{Poisson}(\lambda p)
    hi, with reference to the above posted equation, how did u get from:
    = e^{-\lambda} \frac {(\lambda p)^y} {y!} \sum_{k=y}^{+\infty}
\frac {[\lambda (1 - p)]^{k-y}} {(k - y)!}
    to
    = e^{-\lambda} \frac {(\lambda p)^y} {y!} \sum_{k=0}^{+\infty}
\frac {[\lambda (1 - p)]^{k}} {k!}

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    Re: conditional binomial is a poisson distribution

    Quote Originally Posted by kestertan View Post
    hi, with reference to the above posted equation, how did u get from:
    = e^{-\lambda} \frac {(\lambda p)^y} {y!} \sum_{k=y}^{+\infty}\frac {[\lambda (1 - p)]^{k-y}} {(k - y)!}
    to
    = e^{-\lambda} \frac {(\lambda p)^y} {y!} \sum_{k=0}^{+\infty}\frac {[\lambda (1 - p)]^{k}} {k!}
    Instead of indexing by k what happens if you index using something like j = k-y?
    I don't have emotions and sometimes that makes me very sad.

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    Re: conditional binomial is a poisson distribution

    Quote Originally Posted by Dason View Post
    Instead of indexing by k what happens if you index using something like j = k-y?
    oh so are we able to just index the number to just j = k-y? are there any conditions in order to be able to index the number as such?
    sorry for immediately understanding it, im quite new to probability.

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    Re: conditional binomial is a poisson distribution

    **
    Quote Originally Posted by kestertan View Post
    oh so are we able to just index the number to just j = k-y? are there any conditions in order to be able to index the number as such?
    sorry for not immediately understanding it, im quite new to probability.
    sorry made a typo in the previous message thanks for replying btw

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    Re: conditional binomial is a poisson distribution

    Quote Originally Posted by kestertan View Post
    **

    sorry made a typo in the previous message thanks for replying btw
    ok i have figured that out, using euler formula to solve the equation thanks for all the help!

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    Re: conditional binomial is a poisson distribution


    Hello all

    What will happen if X takes 0


    Quote Originally Posted by BGM View Post
    For u = 0, 1, 2, ...,

    \Pr\{X - Y = u\} 
= \sum_{k=u}^{+\infty} \Pr\{X - Y = u|X = k\}\Pr\{X = k\}

    =\sum_{k=u}^{+\infty} \Pr\{Y = k - u|X = k\}\Pr\{X = k\}

    =\sum_{k=u}^{+\infty} {k \choose k - u}p^{k-u}(1 - p)^u 
e^{-\lambda} \frac {\lambda^k} {k!}

    = e^{-\lambda} \frac {[\lambda(1 - p)]^u} {u!} 
\sum_{k=u}^{+\infty} \frac {(\lambda p)^{k-u}} {(k -u)!}

    = e^{-\lambda} \frac {[\lambda(1 - p)]^u} {u!}
\sum_{k=0}^{+\infty} \frac {(\lambda p)^k} {k!}

    = e^{-\lambda} \frac {[\lambda(1 - p)]^u} {u!} e^{\lambda p}
= e^{-\lambda(1 - p)} \frac {[\lambda(1 - p)]^u} {u!}

    \Rightarrow X - Y \sim \mathrm{Poisson}(\lambda(1 - p))

    Then consider

    \Pr\{Y = y, X - Y = u\} 
= \Pr\{Y = y, X - Y = u|X = u + y\}\Pr\{X = u + y\}

    = \Pr\{Y = y|X = u + y\}\Pr\{X = u + y\}

    = {u + y \choose y } p^y (1-p)^u e^{-\lambda} \frac {\lambda^{u+y}} {(u+y)!}

    = e^{-\lambda p} \frac {(\lambda p)^y} {y!}
e^{-\lambda(1 - p)} \frac {[\lambda (1-p)]^u} {u!}

    = \Pr\{Y = y\}\Pr\{X - Y = u\} ~~ \forall u, y

    \Rightarrow Y, X - Y are independent.

    Finally, \Pr\{X = k|Y = y\} = \Pr\{X - y = k - y|Y = y\}
= \Pr\{X - Y = k - y|Y = y\}

    = \Pr\{X - Y = k - y\} 
= e^{-\lambda(1 - p)} \frac {[\lambda(1 - p)]^{k - y}} {(k-y)!}, 
k = y, y+1, y+2, ...

    which is a Poisson distribution shifted by y

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